CS110: Models of Computing

Lecture 5V. Kamakoti

10th January 2008

• Number Systems and conversions• Decimal• Binary• Octal• Hexadecimal

• Representation of signed integers inBinary• Two’s complement representation

• Floating point in Binary system• IEEE 754 - later in this course

Today’s Lecture Contents

Decimal System

• 986 = 9 * 102 + 8 * 101 + 6 * 100

• The digits used are 0,1,2,…,9• Adding and subtracting numbers in

decimal is well known.• Some interesting facts about

multiplication and division are asfollows.

Decimal Mult and Div

• Multiply by 10 is equivalent to Left shift thedigits by one position and appending a zero

• For eg. 986 * 10 = 9860• Dividing by 10 is to right shift the digits by

one. What remains is the quotient and getsout is the reminder.

• 986/10, gives 98 as quotient and 6 asreminder.

More interesting facts

• Multiplying a number by 10k is to left shift thenumber k times appending zeros.

• 986 * 103 = 986000• Dividing a number by 10k is to right shift it by

k times. What remains is the quotient andleaves is the reminder.

• 98657 / 104 gives 9 as quotient and 8657 asreminder.

• True for any base “b”.

The Binary system• Digits used are 0 and 1• Best suited for computers• (1110)2 = 1*23 + 1*22 + 1*21 + 1*20 = (14)10• Multiplying a binary number by 2k is equivalent

to left shifting the same k times appending zeros• (1110)2 * 23 = (1110000)2 =• 1*26 + 1*25 + 1*24 + 0*23 + 0*22 + 0*21 +0*20

= (112)10• This also teaches you conversion from binary to

decimal

Any base to decimal conversion

• (1345)6 = 1*63 + 3*62 + 4*61 + 6*60 = (354)10

• Systems with base k greater than 10 usesdigits (0,1,2,3,…,k-1). To avoid confusion 10is represented by A, 11 by B and so on.

• For eg. The hexadecimal (base 16) systemuses the alphabets A to F for representing the“digits” 10 to 15 in order.

The Binary system

• Dividing a binary number by 2k is equivalent to rightshifting the same k times.

• (1110)2 / 23 gives (1)2 as quotient and (110)2

• (14)10/(8)10 gives (1)10 as quotient and (6)10 as reminder.• The rightmost bit is called the Least Significant Bit (LSB)

and the leftmost bit is called the Most Significant Bit(MSB).

• For eg. in (1110)2, 1 is the MSB and 0 is the LSB.

Interesting facts• If a binary number is divided by 2, the least

significant bit is the reminder. The next toleast significant bit becomes the lsb (rightshifting).

• 1110 / 2 gives 111 as Q and 0 as reminder• 111/2 gives 11 as Q and 1 as reminder• 11/2 gives 1 as Q and 1 as reminder• 1/2 gives 0 as Q and 1 as reminder• Read the reminders from bottom to top and

that is the original number.

Decimal to Binary

• Repeatedly divide the given number by 2 tillthe quotient becomes zero

• Read out the remainders in the reverse order.• 14/2 = 7 (Quo) and 0 (Rem)• 7/2 = 3 (Quo) and 1 (Rem)• 3/2 = 1 (Quo) and 1 (Rem)• 1/2 = 0 (Quo) and 1 (Rem)• Therefore (14)10 = (1110)2

Octal System• Base 8 and digits are 0,1,2,3,4,5,6,7.• (756)8 = 7 * 82 + 5 * 81 + 6*80 = (494)10• As 8 = 23,• (756)8 = (111)2*26 +(101)2*23+(110)2*20

• = (111 101 110)2• Replace every digit in octal by equivalent 3-bit

binary to convert octal to binary.• Do vice versa for binary to octal• So this is very useful for number

representation in binary computers.

Hexadecimal System• Base 16 and digits are

– 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F.• (AB6)16 = 10 * 162 + 11 * 161 + 6*160 = (2742)10• As 16 = 24,• (AB6)16 = (1010)2*28 +(1011)2*24+(0110)2*20

• = (1010 1011 0110)2• Replace every digit in hexa by equivalent 4-bit binary

to convert hexa to binary.• Do vice versa for binary to hexa• So this is also very useful for number representation

in binary computers.

Signed Integer Representation• We cannot store integers of any size• There is limited space allotted by the system

to store a number - say “n” bits.• So if the number you try to store exceeds 2n -

1, then only the first n-bits from right arestored. The most significant bits are lost.

• So let “A” be any number represented using“n” bits, then in our computer arithmetic,2n + A = A.

• Given n = 4 and A = 11, 16 + 11 = 27, whichis (11011)2. Since only 4 bits forrepresentation, we get (1011)2 which is 11.

Signed Integers• Given A, define Ac as the number got by

toggling every bit.• Given A = 1010, then Ac = 0101.• If A is represented using n bits, then

– A + Ac = 2n - 1.– Above example, n = 4, 1010 + 0101 = 1111 = 15.

• Ac + 1 = 2n - A = 2n + (-A) = -A• To get negative of a number take its positive

representation, toggle (complement) every bitand add one. Hence, named 2’s complement.

Signed Integers• n = 3• 0 = (000)• 1 = (001); -1 = 110 + 1 = (111)• 2 = (010); -2 = 101 + 1 = (110)• 3 = (011); -3 = 100 + 1 = (101)• 4 = (100); -4 = 011 + 1 = (100)• So we represent -4,-3,-2,-1,0,1,2,3• Note that the MSB is 1 for negative and 0 for positive

numbers. Removed +0 and -0 problem.• Easy to see that with n bits the range is -2n-1 to 2n-1 -1

The Circle of Numbers000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -4

Addition Clockwise (Eg. 1+2)000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -4

Addition Clockwise (Eg. -3+2)000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -4

Addition Clockwise (Eg. -2+3)000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -4

Addition Clockwise (Eg. 2+3)000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -4Overflow!! Adding two +ve numbers result in a -ve number

3-bits range:

-4 to +3 only.

Subtraction Anti Clockwise000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -41 - 3; start at 1 and move anticlockwise 3 steps

Subtraction Anti Clockwise000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -41 - 3; start at 1 and move clockwise 5 steps. -3 is related to 5!!

Subtraction Anti Clockwise000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -45 is two’s complement of 3

Subtraction Anti Clockwise000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -4Subtraction is addition!!!!, A - B = A + Two’s complement(B)

Subtraction Anti Clockwise000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -4True for other numbers - Sub by 2 = add by 6

Subtraction Anti Clockwise000 = 0

001 = 1

010 = 2

011 = 3

111 = -1

110 = -2

101 = -3

100 = -4-2 -3 is -2 + 5 = 3; underflow. With 3 bits -4 to 3 only.

Conclusions

• Given that the two input numbers arerepresentable using “n” bits– Overflow occurs only when adding two +ve

numbers– Underflow occurs only when adding two

negative numbers.

Test Your Understanding

• Convert to binary (27)10, (CAFÉ)16, (653)8• Convert to Decimal (56A)13• Convert to Octal, Decimal and Hexadecimal

– (111001110111011)2

• Assuming inputs are correctly representedperform the following– 3 - 2; 5 - 6; 4 + 3; -4 -3; 2 + 7;– All numbers are in decimal above. Indicate

Underflow and overflows

Answers

Creative Question - 3

• You are given a BLACK BOX - nothing isknown what is inside it - but it does thefollowing for you– Give a sequence of n real numbers and an integer

k, the box will output “yes” or “no” indicatingwhether there is a subset of the numbers whosesum is exactly “k”. It will not tell you the subset.

– Use this black box to find the subset whose sum is“k”, if it exists.

– You are allowed to use the black box exactly “n”times, where “n” is the length of the sequence.

Example

(1.3,.7,2.5,1.7) and k = 3

Yes

The problem is using the above box exactly four times you

Should find out the subset that adds to 3, namely (1.3,1.7)

Thank You